2121407 Semigroups. - Harvard Universitypeople.math.harvard.edu/~shlomo/212a/07.pdf · The power series expansion of etA when A is bounded.Unbounded operators, their resolvents and

The power series expansion of etA when A is bounded. Unbounded operators, their resolvents and their spectra. Sectorial operators.

2121407Semigroups.

Shlomo Sternberg

September 23, 2014

Shlomo Sternberg

2121407 Semigroups.


Reminder:

No class this Thursday.

Shlomo Sternberg

2121407 Semigroups.


The semi-group generated by an operator

In today’s lecture I want to discuss the semi-group generated by anoperator A, that is the semi-group

t 7→ etA.

I will start with the simplest case of a bounded operator, and thengeneralize.

Over the course of this and the next two lectures I hope to coverthe Hille-Yosida theorem on semigroups, derive Stone’s theorem onunitary one parameter groups on a Hilbert space (a theorem whichlies at the foundations of quantum mechanics) and the spectraltheorem for (possibly) unbounded self-adjoint operators on aHilbert space.

The key tool that we will use many times in today’s lecture is theCauchy integral formula from elementary complex analysis.

Shlomo Sternberg

2121407 Semigroups.


1 The power series expansion of etA when A is bounded.The resolvent from the semi-group.The semigroup from the resolvent.The two resolvent identities.

2 Unbounded operators, their resolvents and their spectra.

3 Sectorial operators.Definition of a sectorial operator.Definition of etA when A is sectorial.The semi-group property.Bounds on etA.The derivatives of etA, its holomorphic charcter.The limit of etA as t → 0+.The resolvent as the Laplace transform of the semigroup.

Shlomo Sternberg

2121407 Semigroups.


etA when A is bounded

Suppose that A is a bounded operator on a Banach space. Forexample, any linear operator on a finite dimensional space. Thenthe series

etA =∞∑

0

tk

k!Ak

converges for any t. (We will concentrate on t real, and eventuallyon t ≥ 0 when we get to more general cases.) Convergence isguaranteed as a result of the convergence of the usual exponentialseries in one variable. (There are serious problems with thisdefinition from the point of view of numerical implementationwhich we will not discuss here.)

Shlomo Sternberg

2121407 Semigroups.


Identities of etA when A is bounded

The standard proof using the binomial formula shows that

e(s+t)A = esA · etA.

Also, the standard proof for the usual exponential series shows thatthe operator valued function t 7→ etA is differentiable (in theuniform topology) and that

d

dt

(etA)

= A · etA = etA · A.

Shlomo Sternberg

2121407 Semigroups.


The resolvent set and the resolvent.

A point z ∈ C is said to belong to the resolvent set of A if theoperator zI − A has a bounded (two sided) inverse. Then thisinverse is called the resolvent of A and is denoted by R(z ,A). So

Shlomo Sternberg

2121407 Semigroups.


The series for the resolvent when |z | > ‖A‖

R(z ,A) := (zI − A)−1.

For example, if z 6= 0 we have zI − A = z(I − z−1A). If |z | > ‖A‖the geometric series

I + z−1A + z−2A2 + z−3A3 + · · ·

converges to (I − z−1A)−1. So all z satisfying |z | > ‖A‖ belong tothe resolvent set of A and for such z we have the convergent seriesexpansion

R(z ,A) = z−1I + z−2A + z−3A2 + z−4A3 + · · · .

Shlomo Sternberg

2121407 Semigroups.


The spectrum

Of course, in finite dimensions, λ ∈ C is by definition an eigenvalueof A if and only if the operator λI − A is not invertible. So in finitedimensions the resolvent set of A consists of all complex numberswhich are not eigenvalues.

In general, we will define the spectrum to be the complement ofthe resolvent set.

Shlomo Sternberg

2121407 Semigroups.


The resolvent from the semi-group.


Suppose that Re z > ‖A‖ so that z belongs to the resolvent set ofA and also the function

t 7→ e−ztetA

is integrable over (0,∞). We have

(zI − A)

∫ ∞

0e−ztetAdt =

∫ ∞

0(zI − A)e−t(zI−A)dt

= −∫ ∞

0

d

dt

(e−t(zI−A)

)dt = I .

SoShlomo Sternberg

2121407 Semigroups.



R(z ,A) =

∫ ∞

0e−ztetAdt.

In words: for Re z > ‖A‖ the resolvent is the Laplace transformof the one parameter group.

If we differentiate both sides of the above equation n times withrespect to z we find that

R(z ,A)n+1 =1

n!

∫ ∞

0tne−ztetAdt.

Shlomo Sternberg

2121407 Semigroups.


The semigroup from the resolvent.


Let Γ be a circle of radius > ‖A‖ centered at the origin. The claimis that

etA =1

2πi

∫

ΓetλR(λ,A)dλ.

Shlomo Sternberg

2121407 Semigroups.



Proof.

From the power series expansion of the resolvent, the contourintegral is

1

2πi

∞∑

n=0

tn

n!

∞∑

k=0

∫

Γλn

Ak

λk+1dλ =

∞∑

n=0

tn

n!

∞∑

k=0

Ak 1

2πi

∫

Γλn−k−1dλ.

By the Cauchy integral formula, 12πi

∫Γ λ

n−k−1dλ = 0 unless n = kin which case the integral is 1. So the above expression becomes

∞∑

n=0

tnAn

n!= eAt .

Shlomo Sternberg

2121407 Semigroups.



If we integrate the equation

etA =1

2πi

∫

ΓetλR(λ,A)dλ

by parts we obtain

tetA =1

2πi

∫

ΓetλR(λ,A)2dλ.

More generally, integrating by parts n times gives

tn

n!etA =

1

2πi

∫

ΓetλR(λ,A)n+1dλ.

Shlomo Sternberg

2121407 Semigroups.


The two resolvent identities.

The first resolvent identity.

If z and w both belong to the resolvent set of A, then we canmultiply the equation

zI − A = wI − A + (z − w)I

on the left by R(z ,A) and on the right by R(w ,A) to obtain

R(w ,A) = R(z ,A) + (z − w)R(z ,A)R(w ,A)

which is known as the first resolvent identity and dates back tothe 19th century. We will make much use of this identity.

Shlomo Sternberg

2121407 Semigroups.



The second resolvent identity.

The first resolvent identity relates the resolvents of a singleoperator A at two different points in its resolvent set. The secondresolvent identity, which also dates back to the 19th centuryrelates the resolvents of two different operators at a point whichbelongs to the resolvent set of both:

Shlomo Sternberg

2121407 Semigroups.



Let φ and ψ be invertible operators. Clearly

ψ = φ+ ψ(φ−1 − ψ−1)φ.

Let A and B be two operators (at the moment both bounded) andlet z belong to the resolvent set of both A and B. Apply the aboveequation to φ = R(z ,A) and ψ = R(z ,B) so as to get the secondresolvent identity

R(z ,B) = R(z ,A) + R(z ,B)(B − A)R(z ,A).

Shlomo Sternberg

2121407 Semigroups.



eA · eB 6= eA+B in general.

If A and B are bounded operators, then it is not true in generalthat eA · eB = eA+B . Indeed, the two sides agree up to termswhich are linear in A and B, but the quadratic terms on the leftare 1

2

[A2 + 2AB + B2

]while the quadratic terms on the right are

12

[A2 + AB + BA + B2

]. These do not agree unless AB = BA.

So it is not true in general that

eB = e(B−A) · eA.

Indeed, the Campbell-Baker-Hausdorff formula gives a rathercomplicated formula for the operator C such that eB = eC · eA.

However consider the following idea of Kantorovitz:

Shlomo Sternberg

2121407 Semigroups.



Define

X0 := I , X1 = B − A, X2 := B2 − 2BA + A2,

and, in general,

Xn := Bn − nBn−1A +

(n2

)Bn−2A2 + · · · ± An. (1)

In other words, Xn looks like the binomial expansion of (B − A)n

with all the B’s moved to the left and all the A’s to the right.Then, claim:

etB =

(I + tX1 +

1

2t2X2 + · · ·

)etA. (2)

Shlomo Sternberg

2121407 Semigroups.



To prove:

etB =

(I + tX1 +

1

2t2X2 + · · ·

)etA. (2)

Proof.

If A and B commute, this is simply the assertion thatetB = et(B−A)etA. But in trying to verify (2) all the A’s lie to theright of all the B’s, and we never move an A past a B, so (2) istrue in general.

Shlomo Sternberg

2121407 Semigroups.


Unbounded operators

Up until now we have dealt with bounded operators. But we areinterested in partial differential operators such as the heat operatorwhich (at least when acting on a fixed Banach space) areunbounded. So we must discuss unbounded operators.

We will find that for certain types of operators (sectorial operators- see later for the definition) the above discussion about thesemi-group generated by an operator goes through with minormodification in the statements and a good bit of work in theproofs. But for more general unbounded operators (as in theHille-Yosida theorem, which I plan to discuss later) we will have todo major reworking.

Shlomo Sternberg

2121407 Semigroups.


The direct sum of two Banach spaces

Let B and C be Banach spaces. We make B ⊕ C into a Banachspace via

‖{x , y}‖ = ‖x‖+ ‖y‖.Here we are using {x , y} to denote the ordered pair of elementsx ∈ B and y ∈ C so as to avoid any conflict with our notation forscalar product in a Hilbert space. So {x , y} is just another way ofwriting x ⊕ y .

Shlomo Sternberg

2121407 Semigroups.


Linear operators and their graphs.

A subspaceΓ ⊂ B ⊕ C

will be called a graph (more precisely a graph of a lineartransformation) if

{0, y} ∈ Γ ⇒ y = 0.

Another way of saying the same thing is

{x , y1} ∈ Γ and {x , y2} ∈ Γ ⇒ y1 = y2.

In other words, if {x , y} ∈ Γ then y is determined by x .Shlomo Sternberg

2121407 Semigroups.


The domain and the map of a graph.

Let D(Γ) denote the set of all

x ∈ B such that there is a y ∈ C with {x , y} ∈ Γ.

Then D(Γ) is a linear subspace of B, but, and this is veryimportant, D(Γ) is not necessarily a closed subspace. We have alinear map

T (Γ) : D(Γ)→ C , Tx = y where {x , y} ∈ Γ.

Shlomo Sternberg

2121407 Semigroups.


The graph of a linear transformation.

Equally well, we could start with the linear transformation:Suppose we are given a (not necessarily closed) subspaceD(T ) ⊂ B and a linear transformation

T : D(T )→ C .

We can then consider its graph Γ(T ) ⊂ B ⊕ C which consists of all

{x ,Tx}, x ∈ D(T ).

Shlomo Sternberg

2121407 Semigroups.


Thus the notion of a graph, and the notion of a lineartransformation defined only on a subspace of B are logicallyequivalent. When we start with T (as usually will be the case) wewill write D(T ) for the domain of T and Γ(T ) for thecorresponding graph.

There is a certain amount of abuse of language here, in that whenwe write T , we mean to include D(T ) and hence Γ(T ) as part ofthe definition.

Shlomo Sternberg

2121407 Semigroups.


Closed linear transformations.

A linear transformation is said to be closed if its graph is a closedsubspace of B ⊕ C .

Let us disentangle what this says for the operator T . It says that iffn ∈ D(T ) then

fn → f and Tfn → g ⇒ f ∈ D(T ) and Tf = g .

This is a much weaker requirement than continuity. Continuity ofT would say that fn → f alone would imply that Tfn converges toTf . Closedness says that if we know that both

fn converges and gn = Tfn converges to g

then f = lim fn lies in D(T ) and Tf = g .Shlomo Sternberg

2121407 Semigroups.


Avoiding the closed graph theorem

It is here that we run up against a famous theorem - the closedgraph theorem - which says that if T is defined on all of a Banachspace B and has a closed graph, then T must be bounded.

So if we are considering operators which are not bounded, we haveto deal with operators whose domain is not all of B.

Shlomo Sternberg

2121407 Semigroups.


The resolvent, the resolvent set, and the specturm.

Let T : B → B be an operator with domain D = D(T ). Acomplex number z is said to belong to the resolvent set of T ifthe operator

zI − T

maps D onto all of B and has a two sided bounded inverse. Asbefore, we denote this bounded inverse by R(z ,T ) or Rz(T ) orsimply by Rz if T is understood. So

R(z ,T ) := (zI − T )−1 maps B → D(T )

and is bounded. R(z ,T ) is called the resolvent of T at thecomplex number z . The complement of the resolvent set is calledthe spectrum of T and is denoted by Spec(T ).

Shlomo Sternberg

2121407 Semigroups.


The spectrum is a closed subset of C.

Theorem

The set Spec(T ) is a closed subset of C. In fact, if z 6∈ Spec(T )and c := ‖R(z ,T )‖ then the spectrum does not intersect the disk

{w ∈ C| |(w − z)| < c−1}.

For w in this disk R(w ,T ) =∑∞

0 (−(w − z))nR(z ,T )n+1 andso is an analytic operator valued function of w. Differentiating thisseries term by term shows that

d

dzR(z ,T ) = −R(z ,T )2.

Shlomo Sternberg

2121407 Semigroups.


Proof, part 1.

The series given in the theorem certainly converges in operatornorm to a bounded operator for w in the disk. For a fixed w in thedisk, let C denote the operator which is the sum of the series.Then

C = R(z ,T )− (w − z)R(z ,T )C .

This shows that C maps B to D(T ) and has kernel equal to thekernel of R(z ,T ) which is {0}. So C is a bounded injectiveoperator mapping B into D.

Shlomo Sternberg

2121407 Semigroups.


AlsoC = R(z ,T )− (w − z)CR(z ,T )

which shows that the image of R(z ,T ) is contained in the imageof C and so the image of C is all of D.

Shlomo Sternberg

2121407 Semigroups.


Proof, part 2.

C :=∞∑

0

(−(w − z))nR(z ,T )n+1.

If f ∈ D and g = (zI − T )f then f = R(z ,T )g and soCg = f − (w − z)Cf and hence

C (zf − Tf ) = f − (w − z)Cf

orC (−Tf ) = f − wCf so C (wI − T )f = f

showing that C is a left inverse for wI − T .

Shlomo Sternberg

2121407 Semigroups.


A similar argument shows that it is a right inverse. So we haveproved that the series converges to the resolvent proving that theresolvent set is open and hence that the spectrum is closed. Therest of the theorem is immediate. �

The two resolvent identities follow just as in the bounded case.

Shlomo Sternberg

2121407 Semigroups.


Lemma

If T : B → B is an operator on a Banach space whose spectrum isnot the entire plane then T is closed.

Shlomo Sternberg

2121407 Semigroups.


Proof.

Assume that R = R(z ,T ) exists for some z . Suppose that fn is asequence of elements in the domain of T with fn → f andTfn → g . Set hn := (zI − T )fn so

hn → zf − g .

Then R(zf − g) = lim Rhn = lim fn = f . Since R maps B to thedomain of T this shows that f lies in this domain. MultiplyingR(zf − g) = f by zI − T gives

zf − g = zf − Tf

showing that Tf = g .

Shlomo Sternberg

2121407 Semigroups.


The adjoint of a densely defined linear operator.

Suppose that we have a linear operator T : D(T )→ C and let usmake the hypothesis that

D(T ) is dense in B.

Any element of B∗ is then completely determined by its restrictionto D(T ). Now consider

Γ(T )∗ ⊂ C ∗ ⊕ B∗

defined by

{`,m} ∈ Γ(T )∗ ⇔ 〈`,Tx〉 = 〈m, x〉 ∀ x ∈ D(T ). (3)

Shlomo Sternberg

2121407 Semigroups.


The adjoint of a densely defined linear operator.

〈`,Tx〉 = 〈m, x〉 ∀ x ∈ D(T ). (3)

Since m is determined by its restriction to D(T ), we see thatΓ∗ = Γ(T ∗) is indeed a graph. (It is easy to check that it is alinear subspace of C ∗ ⊕ B∗.) In other words we have defined alinear transformation

T ∗ := T (Γ(T )∗)

whose domain consists of all ` ∈ C ∗ such that there exists anm ∈ B∗ for which 〈`,Tx〉 = 〈m, x〉 ∀ x ∈ D(T ).

Shlomo Sternberg

2121407 Semigroups.


The adjoint of a linear transformation is closed.

If `n → ` and mn → m then the definition of convergence in thesespaces implies that for any x ∈ D(T ) we have

〈`,Tx〉 = lim〈`n,Tx〉 = lim〈mn, x〉 = 〈m, x〉.

If we let x range over all of D(T ) we conclude that Γ∗ is a closedsubspace of C ∗ ⊕ B∗. In other words we have proved

Theorem

If T : D(T )→ C is a linear transformation whose domain D(T ) isdense in B, it has a well defined adjoint T ∗ whose graph is givenby (3). Furthermore T ∗ is a closed operator.

Shlomo Sternberg

2121407 Semigroups.


Most of the material for the rest of today’s lecture (including thefigures) is taken from:

Analytic Semigroups and Reaction-Diffusion ProblemsInternet Seminar 2004 - 2005

byLuca Lorenzi, Alessandra Lunardi, Giorgio Metafune, Diego Pallara

and from

Introduzione ai problemi parabolici nonlineariby

Alessandra Lunardi

via the internet.

Shlomo Sternberg

2121407 Semigroups.


Definition of a sectorial operator.

Sectorial operators.

A closed operator A is called sectorial if its resolvent set containsa sector S of the form

S = {λ ∈ C|λ 6= 0, | arg λ| < θ, θ > π/2}.

More precisely, we will assume that there is a positive constant Msuch that

‖(λI − A)−1‖ ≤ M

|λ| , λ ∈ S .

For example, we have seen that the spectrum of the LaplacianA = −∆ on a compact manifold is discrete, lies along the negativereal axis and tends to −∞. Hence A is sectorial.

Shlomo Sternberg

2121407 Semigroups.


Definition of etA when A is sectorial.


Recall that in the case of a bounded operator A, the exponentialseries for etA converges, and if Γ is a circle of radius > ‖A‖centered at the origin, we proved that

etA =1

2πi

∫

ΓetλR(λ,A)dλ.

For sectorial operators we will define etA by a contour integral ofetλR(λ,A) over a modified contour. I think that the idea of doingthis goes back to Dunford.

Here is the picture of the proposed contour, where we have“opened up” the circle by two branches going off diagonally to theleft:

Shlomo Sternberg

2121407 Semigroups.



10 Chapter 1

1.3 Sectorial operators

Definition 1.3.1 We say that a linear operator A : D(A) ! X " X is sectorial if thereare constants ! # R, " # (#/2,#), M > 0 such that

!""#""$

(i) $(A) $ S!," := {% # C : % %= !, | arg(%& !)| < "},

(ii) 'R(%, A)'L(X) (M

|%& !| , % # S!,".(1.9)

Note that every sectorial operator is closed, because its resolvent set is not empty.

For every t > 0, the conditions (1.9) allow us to define a bounded linear operator etA

on X, through an integral formula that generalizes (1.6). For r > 0, & # (#/2, "), let 'r,#

be the curve

{% # C : | arg %| = &, |%| ) r} *{ % # C : | arg %| ( &, |%| = r},

oriented counterclockwise, as in Figure 1.

&

!

'r,! + !

! + r

((A)

Figure 1.1: the curve 'r,#.

For each t > 0 set

etA =1

2#i

%

$r,!+"et%R(%, A) d%, t > 0. (1.10)

Using the obvious parametrization of 'r,# we get

etA =e"t

2#i

&&% +!

re(& cos #"i& sin #)tR(! + $e"i#, A)e"i#d$

+

% #

"#e(r cos'+ir sin')tR(! + rei', A)irei'd)

+

% +!

re(& cos #+i& sin #)tR(! + $ei#, A)ei#d$

',

(1.11)

for every t > 0 and for every r > 0, & # (#/2, ").

Shlomo Sternberg

2121407 Semigroups.



In the figure, we have slightly generalized the notion of a sectorialoperator to allow for some positive spectrum, so we assume thatthe resolvent set contains the set

Sθ,ω := {λ ∈ C|λ 6= ω, | arg(λ− ω)| < θ}

where ω ∈ R and θ ∈ (π/2, π). We also assume that

‖R(λ,A‖ ≤ M

|λ− ω| λ ∈ Sθ,ω

for some positive constant M. We let γr ,η be the curve

{λ ∈ C| | arg(λ)| = η, |λ| ≥ r} ∪ {λ ∈ C| | arg λ| ≤ η, |λ| = r}.

The curve in the figure is the curve γr ,η shifted by ω.

Shlomo Sternberg

2121407 Semigroups.



10 Chapter 1



!""#""$

(i) $(A) $ S!," := {% # C : % %= !, | arg(%& !)| < "},

(ii) 'R(%, A)'L(X) (M

|%& !| , % # S!,".(1.9)




be the curve

{% # C : | arg %| = &, |%| ) r} *{ % # C : | arg %| ( &, |%| = r},


&

!

'r,! + !

! + r

((A)


For each t > 0 set

etA =1

2#i

%

$r,!+"et%R(%, A) d%, t > 0. (1.10)


etA =e"t

2#i

&&% +!


+

% #


+

% +!


',

(1.11)


For each t > 0 we define

etA :=1

2πi

∫

γr,η+ωetλR(λ,A)dλ.

Shlomo Sternberg

2121407 Semigroups.



We must show that:

This is well defined, i.e. that the integral converges anddefines a bounded operator.

This has the properties we expect of the exponential, e.g.esAetA = e(s+t)A, and AketA = etAAk (when defined).

The definition is independent of the choice of r and η.

This coincides with the old definition when A is bounded.

Shlomo Sternberg

2121407 Semigroups.



The integral converges.

10 Chapter 1



!""#""$

(i) $(A) $ S!," := {% # C : % %= !, | arg(%& !)| < "},

(ii) 'R(%, A)'L(X) (M

|%& !| , % # S!,".(1.9)




be the curve

{% # C : | arg %| = &, |%| ) r} *{ % # C : | arg %| ( &, |%| = r},


&

!

'r,! + !

! + r

((A)


For each t > 0 set

etA =1

2#i

%

$r,!+"et%R(%, A) d%, t > 0. (1.10)


etA =e"t

2#i

&&% +!


+

% #


+

% +!


',

(1.11)


The integral breaks up into three pieces given by

eωt

2πi

(−∫ ∞

re(ρ cos η−iρ sin η)tR(ω + ρe−iη,A)e−iηdρ

+

∫ η

−ηe(r cosα+ir sinα)tR(ω + re iα,A)ire iαdα

+

∫ ∞

re(ρ cos η+iρ sin η)tR(ω + ρe iη,A)e iηdρ

).

Shlomo Sternberg

2121407 Semigroups.



The middle integral causes no convergence problems. For the firstand third terms, we know that ‖R(ω + ρe±iη,A)‖ is bounded byM/r and cos η < 0 so there is no trouble with these integrals.

We now show that the integral is independent of the choice of rand η: For this choose a different r ′, η′ and consider the region Dlying between the two curves γr ,η + ω and γr ′η′ + ω lying betweenthem and the cutoff regions Dn := D ∩ {|z − ω| < n} for n ∈ N.See the figure on the next slide.

Shlomo Sternberg

2121407 Semigroups.



1.3. Sectorial operators 11

Lemma 1.3.2 If A is a sectorial operator, the integral in (1.10) is well defined, and it isindependent of r > 0 and ! ! ("/2, #).

Proof. First of all, notice that for each t > 0 the mapping $ "# et!R($, A) is a L(X)-valued holomorphic function in the sector S",# (see Proposition B.4). Moreover, for any$ = % + rei", the estimate

$et!R($, A)$L(X) % exp(%t) exp(tr cos !)M

r(1.12)

holds for each $ in the two half-lines, and this easily implies that the improper integral isconvergent. Now take any r! > 0, !! ! ("/2, #) and consider the integral on &r!,$! + %. LetD be the region lying between the curves &r,$ + % and &r!,$! + % and for every n ! N setDn = D & {|z ' %| % n}, as in Figure 1.2. By Cauchy integral theorem A.9 we have

!

%Dn

et!R($, A) d$ = 0.

By estimate (1.12), the integrals on the two arcs contained in {|z ' %| = n} tend to 0 asn tends to +(, so that

!

&r,!+#et!R($, A) d$ =

!

&r!,!!+#et!R($, A) d$

and the proof is complete. !

%

n

&r,! + %

&r!,!! + %

Dn

Figure 1.2: the region Dn.

Let us also sete0Ax = x, x ! X. (1.13)

In the following theorem we summarize the main properties of etA for t > 0.

Shlomo Sternberg

2121407 Semigroups.



The function λ 7→ etλR(λ,A) is holomorphic on the domain Dn

and hence its integral over ∂Dn vanishes. The integral over thepieces of the circle |λ− ω| = n lying on the boundary tend to zeroby the estimate we gave above, namely that over these integralsthe term cosα is negative. This proves that the definition of etA

does not depend on the choice of r and η.

If A is a bounded operator, we choose r > ‖A‖ and let η → π.The same argument shows that the integral tends to the integralover the circle, as the two radial integrals will tend to terms whichcancel one another. So the new definition coincides with the oldone for bounded operators.

We define e0A = I .

Shlomo Sternberg

2121407 Semigroups.



We now prove:

Proposition

For all t > 0, etA : X → Dom(A) and for x ∈ Dom(A)

AetAx = etAAx .

For this purpose, we first prove a lemma about closed operators A:

Shlomo Sternberg

2121407 Semigroups.



Lemma

Let I = (a, b) with −∞ ≤ a < b ≤ +∞ and f : I → Dom(A) be afunction which is Riemann integrable, and such that the functionλ 7→ Af (λ) is also Riemann integrable. Then

∫

If (λ)dλ ∈ Dom(A) and A

∫

If (λ)dλ =

∫

IAf (λ)dλ.

Shlomo Sternberg

2121407 Semigroups.



Proof of the lemma.

First suppose that I is bounded. Then the assertion S ∈ Dom(A)is true by linearity where S is a Riemann approximating sumS =

∑f (mi )(λi − λi−1) as is the assertion

AS =∑

Af (mi )(λi − λi−1).

Shlomo Sternberg

2121407 Semigroups.



Proof of the lemma, 2.

By assumption S →∫I f (λ)dλ and AS →

∫I Af (λ)dλ. The fact

that A is closed then implies that

∫

If (λ)dλ ∈ Dom(A) and A

∫

If (λ)dλ =

∫

IAf (λ)dλ.

The non-bounded case follows by the same argument since theconvergence of the integrals means that∫I = lim

∫bounded intervals . �

Shlomo Sternberg

2121407 Semigroups.



Proof of the proposition.

Replacing A by A− ωI we may assume that ω = 0 to simplify thenotation. Take f (λ) = etλR(λ,A)x in the lemma: the resolventmaps the entire Banach space X into Dom(A) and hence thelemma implies that etAx ∈ Dom(A) for any x ∈ X and that

AetAx =1

2πi

∫

γr,η

etλAR(λ,A)xdλ.

Now by the definition of the resolvent, (λI − A)R(λ,A) = I soAR(λ,A) = λR(λ,A)− I . By Cauchy,

∫γr,η

etλdλ = 0, so

AetAx =1

2πi

∫

γr,η

λetλR(λ,A)xdλ.

Shlomo Sternberg

2121407 Semigroups.



We have shown that

AetAx =1

2πi

∫

γr,η

etλAR(λ,A)xdλ

=1

2πi

∫

γr,η

λetλR(λ,A)xdλ.

For x ∈ Dom(A) we have R(λ,A)Ax = AR(λ,A)x so the firstequation above shows that for x ∈ Dom(A)

AetAx = etAAx . �

We continue with the assumption that t > 0.

Shlomo Sternberg

2121407 Semigroups.



We know that etAx ∈ Dom(A) for any x ∈ X and if x ∈ Dom(A)then AetAx = etAAx = 1

2πi

∫γr,η

λetλR(λ,A)xdλ.

The lemma applied to the function f (λ) := λetλR(λ,A)x tells usthat AetAx belongs to Dom(A). In other words, etAx belongs toDom(A2) and furthermore

A2etAx =1

2πi

∫

γr,η

AλeλtR(λ,A)xdλ.

If x ∈ Dom(A) we can move the A past the R(λ,A) to concludethat

A2etAx = AetAAx .

Shlomo Sternberg

2121407 Semigroups.



If x ∈ Dom(A) we can move the A past the R(λ,A) to concludethat

A2etAx = AetAAx .

If we assume in addition that x ∈ Dom(A2) so that (by definition)Ax ∈ Dom(A) we can apply the proposition to Ax and concludethat

A2etAx = etAA2x .

Shlomo Sternberg

2121407 Semigroups.



Continuing in this way we conclude

Theorem

If t > 0 then etAx ∈ Dom(Ak) for all positive integers k, and allx ∈ X . If x ∈ Dom(Ak) then

AketAx = etAAkx .

Furthermore

AketA =1

2πi

∫

γr,η

λketλR(λ,A)dλ.

Shlomo Sternberg

2121407 Semigroups.


The semi-group property.

The first resolvent identity.

Recall that this says that for λ and µ in the resolvent set of A wehave:

R(λ,A)− R(µ,A) = (µ− λ)R(λ,A)R(µ,A).

Shlomo Sternberg

2121407 Semigroups.



Proof.

Since (µI − A)R(µ,A) = I and (λI − A)R(λ,A) = I we have

R(λ,A) = [µR(µ,A)− AR(µ,A)]R(λ,A)

R(µ,A) = [λR(λ,A)− AR(λ,A)]R(µ,A).

Subtract the second equation from the first and use the fact thatthe resolvents commute to give the first resolvent identity.

Shlomo Sternberg

2121407 Semigroups.



We are going to use the first resolvent identity in the form

R(λ,A)R(µ,A) =R(λ,A)− R(µ,A)

µ− λ if µ 6= λ

together with the Cauchy integral formula to prove that

etAesA = e(t+s)A.

For this purpose we write

esA =1

2πi

∫

γ2r,η′eµsR(µ,A)dµ

etA =1

2πi

∫

γr,η

eλtR(λ,A)dλ

where η′ ∈ (η, π/2).Shlomo Sternberg

2121407 Semigroups.



So esAetA =(

1

2πi

)2 ∫

γr,η

∫

γ2r,η′

1

µ− λeµseλtR(λ,A)dµdλ

−(

1

2πi

)2 ∫

γ2r,η′

∫

γr,η

1

µ− λeµseλtR(µ,A)dλdµ.

(The order of integration is irrelevant due to the exponentialconvergence.) The 2nd integral, where we first integrate over theinterior curve, vanishes by Cauchy’s theorem, as the integrand isholomorphic. See the right hand figure on the next slide. Cauchy’sintegral formula applied to the inner integrand of the first integralwhere we integrate over the outer curve first, gives,

1

2πi

∫

γr,η

eλteλsR(λ,A)dλ = e(s+t)A. �Shlomo Sternberg

2121407 Semigroups.



14 Chapter 1

The equalitydk

dtketA = AketA, t > 0

can be proved by the same argument, or by recurrence. Now, let 0 < ! < " ! #/2 begiven, and set $ = " ! !. The function

z "# ezA =1

2#i

!

!r,!

ez"R(%, A)d%

is well defined and holomorphic in the sector

S# = {z $ C : z %= 0, | arg z| < " ! #/2 ! !},

because we can di!erentiate with respect to z under the integral, again by Exercise A.6.Indeed, if % = &ei$ and z = 'ei%, then Re(z%) = &' cos($+() & !c&' for a suitable c > 0.

Since the union of the sectors S#, for 0 < ! < " ! #/2, is S&!"2,0, (iv) is proved. !

Statement (ii) in Theorem 1.3.3 tells us that the family of operators etA satisfies thesemigroup law, an algebraic property which is coherent with the exponential notation.Statement (iv) tells us that e·A is analytically extendable to a sector. Therefore, it isnatural to give the following deefinition.

Definition 1.3.4 Let A be a sectorial operator. The function from [0,+') to L(X),t "# etA (see (1.10), (1.13)) is called the analytic semigroup generated by A (in X).

!n

%

)r,!

)2r,!!

µ

)r,!

)2r,!!

!n

Figure 1.3: the curves for Exercise 2.

Exercises 1.3.5

1. Let A : D(A) ( X # X be sectorial, let ! $ C, and set B : D(B) := D(A) # X,Bx = Ax ! !x, C : D(C) = D(A) # X, Cx = !Ax. Prove that the operator B issectorial, and that etB = e!#tetA. Use this result to complete the proof of Theorem1.3.3 in the case * %= 0. For which ! is the operator C sectorial?

2. Prove that (1.18) holds, integrating over the curves shown in Figure 1.3.

Shlomo Sternberg

2121407 Semigroups.


Bounds on etA.

Proposition

There is a constant M0 such that

‖etA‖ ≤ M0eωt .

Once again, for the proof we may assume that ω = 0. Alsoremember that we may replace r by r ′ in the contour over whichwe integrate. In particular, we may replace r by r/t in the contourfor esA, in particular for s = t. We make the change of variablesξ = λt and obtain:

Shlomo Sternberg

2121407 Semigroups.


Bounds on etA.

Proof.


with !! ! (!2 , !), using the resolvent identity it follows that

etAesA =!

12"i

"2 #

"r,!

#

"2r,!!

e#t+µs R(#, A)"R(µ,A)µ" #

d#dµ

=!

12"i

"2 #

"r,!

e#tR(#, A)d#

#

"2r,!!

eµs dµ

µ" #

"!

12"i

"2 #

"2r,!!

eµsR(µ,A)dµ

#

"r,!

e#t d#

µ" #= e(t+s)A,

where we have used the equalities#

"2r,!!

eµs dµ

µ" #= 2" ies#, # ! $r,$,

#

"r,!

e#t d#

µ" #= 0, µ ! $2r,$! (1.18)

that can be easily checked (Exercise 2, §1.3.5).

Proof of (iii). Let us point out that if we estimate #etA# integrating #e#tR(#, A)# over $r,$

we get a singularity near t = 0, because the norm of the integrand behaves like M/|#| for|#| small. We have to be more careful. Setting #t = % in (1.10) and using Lemma 1.3.2,we get

etA =1

2"i

#

"rt,!

e%R

!%

t, A

"d%

t=

12"i

#

"r,!

e%R

!%

t, A

"d%

t

=1

2"i

! # +"

re&ei!

R

!&ei$

t, A

"ei$

td&"

# +"

re&e"i!

R

!&e#i$

t, A

"e#i$

td&

+# $

#$erei"

R

!rei'

t, A

"irei' d'

t

".

It follows that

#etA# $ 1"

$# +"

rMe& cos $ d&

&+

12

# $

#$Mer cos 'd'

%.

The estimate of #AetA# is easier, and we do not need the above procedure. Recalling that#AR(#, A)# $ M + 1 for each # ! $r,$ and using (1.11) we get

#AetA# $ M + 1"

# +"

re&t cos $d& +

(M + 1)r2"

# $

#$ert cos 'd',

so that, letting r % 0,

#AetA# $ M + 1"| cos !|t :=

N

t, t > 0.

From the equality AetAx = etAAx, which is true for each x ! D(A), it follows thatAketA = (Ae

tk A)k for all k ! N, so that

#AketA#L(X) $ (Nkt#1)k := Mkt#k.

Proof of (iv). This follows easily from Exercise A.6 and from (1.17). Indeed,

d

dtetA =

12"i

#

"r,!

#e#tR(#, A)d# = AetA, t > 0.

Shlomo Sternberg

2121407 Semigroups.


Bounds on etA.

Estimating AetA.

Recall that ‖R(λ− ω,A)‖ ≤ M|λ−ω| is part of our assumption, and

that AR(λ,A) = λR(λ,A)− I so on the path γr ,η we have

‖AR(λ,A)‖ ≤ M + 1.

Hence

‖AetA‖ ≤ M + 1

2π

[2

∫ ∞

reρt cos ηdρ+ r

∫ η

−ηert cosαdα

].

Shlomo Sternberg

2121407 Semigroups.


Bounds on etA.

‖AetA‖ ≤ M + 1

2π

[2

∫ ∞

reρt cos ηdρ+ r

∫ η

−ηert cosαdα

].

Let r → 0. The second integral disappears. In the first integral,make the change of variable ρ 7→ s = t| cos η|ρ. We get‖AetA‖ ≤ M+1

πt| cos η|∫∞

0 e−sds so

‖AetA‖ ≤ M + 1

πt| cos η| =:M1

t.

Shlomo Sternberg

2121407 Semigroups.


Bounds on etA.

‖AetA‖ ≤ M + 1

πt| cos η| =:M1

t.

Now etA maps X → Dom(A) and for every x ∈ Dom(A) we haveAetAx = etAAx hence

(AetkA)k = AketA ∀ k ∈ N.

So applying the above inequality we obtain:

‖AketA‖ ≤ Mk

tkwhere Mk := (kM1)k .

Shlomo Sternberg

2121407 Semigroups.


The derivatives of etA, its holomorphic charcter.

The derivatives of etA for t > 0.

I will continue with the harmless assumption that ω = 0.Differentiating the definition

etA :=1

2πi

∫

γr,η

etλR(λ,A)dλ.

under the integral sign we see that for t > 0 we have

d

dtetA :=

1

2πi

∫

γr,η

λetλR(λ,A)dλ,

and this is equal to AetA as we have already seen. So for t > 0 wehave

d

dtetA = AetA.

Shlomo Sternberg

2121407 Semigroups.



Iterating the formulad

dtetA = AetA

givesdk

dtketA = AketA

for t > 0. So t 7→ etA is C∞ as a map from R+ to boundedoperators on X and its derivatives are given as above.

In fact, it extends to a holomorphic function in a wedge about thepositive x-axis as we shall now see.

Shlomo Sternberg

2121407 Semigroups.



The holomorphic character of etA.

The exponent in the integral along the rays in

1

2πi

∫

γr,η

ezλR(λ,A)dλ.

where λ = ρe iη and z = |z |e iτ is

ρ|z |e i(η+τ) = ρ|z |(cos(η + τ) + i sin(η + τ)).

So as long as cos(η+ τ) < 0, which is the same as |τ | < η− π2 this

integral converges, and we can differentiate under the integral sign.So if θ is the angle that enters into the sectorial character of A, wesee that the function z 7→ ezA defined by the above integral forsome |η| < θ is holomorphic as a function of z for arg(z) < θ − π

2 .

Shlomo Sternberg

2121407 Semigroups.


The limit of etA as t → 0+.

The limit of etAx as t → 0 for x ∈ Dom(A).

Let x ∈ Dom(A), choose R 3 ξ > ω so that ξ is in the resolventset of A, and choose 0 < r < ξ − ω. Let y := ξx − Ax so thatx = R(ξ,A)y . Then by the first resolvent identity

Shlomo Sternberg

2121407 Semigroups.




3. Let A : D(A) ! X " X be sectorial and let x # D(A) be an eigenvector of A witheigenvalue !.

(a) Prove that R(µ,A)x = (µ$ !)!1x for any µ # "(A).

(b) Prove that etAx = e!tx for any t > 0.

4. Prove that if both A and $A are sectorial operators in X, then A is bounded.

Given x # X, the function t %" etAx is analytic for t > 0. Let us consider its behaviorfor t close to 0.

Proposition 1.3.6 The following statements hold.

(i) If x # D(A), then limt"0+ etAx = x. Conversely, if y = limt"0+ etAx exists, thenx # D(A) and y = x.

(ii) For every x # X and t & 0, the integral! t0 esAx ds belongs to D(A), and

A

" t

0esAx ds = etAx$ x. (1.19)

If, in addition, the function s %" AesAx is integrable in (0, #) for some # > 0, then

etAx$ x =" t

0AesAx ds, t & 0.

(iii) If x # D(A) and Ax # D(A), then limt"0+(etAx $ x)/t = Ax. Conversely, ifz := limt"0+(etAx$ x)/t exists, then x # D(A) and Ax = z # D(A).

(iv) If x # D(A) and Ax # D(A), then limt"0+ AetAx = Ax.

Proof. Proof of (i). Notice that we cannot let t " 0+ in the Definition (1.10) of etAx,because the estimate 'R(!, A)' ( M/|! $ $| does not su!ce to use any convergencetheorem.

But if x # D(A) things are easier: indeed fix %, r such that $ < % # "(A), 0 < r < %$$,and set y = %x$Ax, so that x = R(%, A)y. We have

etAx = etAR(%, A)y =1

2&i

"

"r,!+#et!R(!, A)R(%, A)y d!

=1

2&i

"

"r,!+#et! R(!, A)

% $ !y d!$ 1

2&i

"

"r,!+#et! R(%, A)

% $ !y d!

=1

2&i

"

"r,!+#et! R(!, A)

% $ !y d!,

because the integral!"r,!+# et!R(%, A)y/(% $ !) d! vanishes (why?). Here we may let

t " 0+ because 'R(!, A)y/(% $ !)' ( C|!|!2 for ! # 'r,$ + $. We get

limt"0+

etAx =1

2&i

"

"r,!+#

R(!, A)% $ !

y d! = R(%, A)y = x.

as the second integral vanishes by the Cauchy integral formula as ξis outside the region to the left of the curve of integration. Theterm R(λ, ξ)/(ξ − λ) in the first integral is of order |λ|−2 forlarge λ so we can pass to the limit t → 0+ under the integral signto obtain

limt→0+

etAx =1

2πi

∫

γr,η+ω

1

ξ − λR(λ,A)ydλ.

Shlomo Sternberg

2121407 Semigroups.



We have shown that

limt→0+

etAx =1

2πi

∫

γr,η+ω

1

ξ − λR(λ,A)ydλ.

Now we can cut off the radial pieces of the integral at |λ− ω| = nand close up the curve via the circular arc |λ− ω| = n going fromarg(λ− ω) = η to arg(λ− ω) = −η. On this circular arc theintegrand is of order |λ|−2 so this portion of the integral goes tozero as n→∞. The point ξ is in the interior of the closed curve.Taking into account that the curve is oriented clockwise ratherthat counter clockwise, the Cauchy integral formula tells us thatthe integral equals R(ξ,A)y = x . So we have proved that

limt→0+

etAx = x

for x ∈ Dom(A).Shlomo Sternberg

2121407 Semigroups.



We have proved thatlimt→0+

etAx = x

for x ∈ Dom(A).

We also know that ‖etA‖ is bounded for 0 < t < 1 by a constantindependent of t, so it follows from the above that

Proposition

If x ∈ Dom(A) thenlimt→0+

etAx = x

Shlomo Sternberg

2121407 Semigroups.



Conversely, suppose that limt→0+ etAx = y . We know thatetAx ∈ Dom(A) so y ∈ Dom(A). For any ξ in the resolvent set ofA we know that R(ξ,A) is a bounded operator which maps all ofX into Dom(A). So

R(ξ,A)y = R(ξ,A) limt→0+

etAx = limt→0+

etAR(ξ,A)x = R(ξ,A)x .

So y = x .

Shlomo Sternberg

2121407 Semigroups.



Proposition

For every x ∈ X and t ≥ 0

1∫ t

0 esAxds ∈ Dom(A) and

2 A∫ t

0 esAxds = etAx − x .

3 If, in addition, the function s 7→ AesAx is integrable on (0, ε)for some ε > 0 then

etAx − x =

∫ t

0AesAxds.

for any t ≥ 0.

Remark. The highlighted step 2 will be very important for us.

Shlomo Sternberg

2121407 Semigroups.



Proof of 1.

Choose ξ in the resolvent set of A. For any ε ∈ (0, t)∫ t

εesAxds =

∫ t

ε(ξI − A)R(ξ,A)esAxds

= ξ

∫ t

εR(ξ,A)esAxds −

∫ t

ε

d

ds[R(ξ,A)esAx ]ds

= ξR(ξ,A)

∫ t

εesAxds − etAR(ξ,A)x + eεAR(ξ,A)x .

Since R(ξ,A)x ∈ Dom(A), the limit of the the last term as ε→ 0is R(ξ,A)x . So∫ t

0esAxds = ξR(ξ,A)

∫ t

0esAxds − etAR(ξ,A)x + R(ξ,A)x .

Shlomo Sternberg

2121407 Semigroups.



Proof of 1, continued, proof of 2 and 3.

We have proved that∫ t

0esAxds = ξR(ξ,A)

∫ t

0esAxds − etAR(ξ,A)x + R(ξ,A)x .

Since R(ξ,A) maps X into Dom(A), we have proved 1.

If we apply ξI − A to both sides of the above equations we get

(ξI − A)

∫ t

0esAxds = ξ

∫ t

0esAxds − (etAx − x)

which after rearranging the terms says thatA∫ t

0 esAxds = etAx − x which is 2. Under the integrabilityhypothesis we may move the A inside the integral which is 3. �

Shlomo Sternberg

2121407 Semigroups.



Proposition

If x ∈ Dom(A) and Ax ∈ Dom(A) then

limt→0+

1

t[etAx − x ] = Ax .

Conversely, if the limit of the left hand side exists, call it z, thenx ∈ Dom(A) and Ax = z,

Proof. By statement 3 of the preceding Proposition, we may writethe left hand side of the equation in our proposition as1t

∫ t0 esAAxdx . Since the function s 7→ esAAx is continuous, we

can apply the fundamental theorem of the calculus to concludethat the limit is indeed Ax .

Shlomo Sternberg

2121407 Semigroups.



Conversely: If the limit exists, then etAx → x so both x and zbelong to Dom(A). For any ξ in the resolvent set of A we have

R(ξ,A)z = limt→0+

R(ξ,A)1

t(etAx − x) = lim

t→0+

1

tR(ξ,A)A

∫ t

0etAxdx

= (ξR(ξ,A)− I ) limt→0+

1

t

∫ t

0etAxdx = ξR(ξ,A)x − x .

This shows that x ∈ Dom(A). If we apply ξI − A to both sides ofthis equation we get

z = ξx − (ξI − A)x

soz = Ax . �

Shlomo Sternberg

2121407 Semigroups.


The resolvent as the Laplace transform of the semigroup.

Recall the formula A∫ t

0 esAxds = etAx − x . Replace A by A− λIfor λ > 0 to obtain

e−λtetAx − x = (A− λI )

∫ t

0e−λsesAxds.

Letting t →∞ gives (since etA is uniformly bounded in t > 0)

x = (λI − A)

∫ ∞

0e−λtetAxdt.

Applying R(λ,A) to this equation gives

R(λ,A) =

∫ ∞

0e−λtetAdt.

This was proved for real positive λ. But by analytic continuationit is true for all λ in the sector S .

Shlomo Sternberg

2121407 Semigroups.


The resolvent as the Laplace transform of the semigroup.

Summary:

1 The power series expansion of etA when A is bounded.The resolvent from the semi-group.The semigroup from the resolvent.The two resolvent identities.

2 Unbounded operators, their resolvents and their spectra.3 Sectorial operators.

Definition of a sectorial operator.Definition of etA when A is sectorial.The semi-group property.Bounds on etA.The derivatives of etA, its holomorphic charcter.The limit of etA as t → 0+.The resolvent as the Laplace transform of the semigroup.

Shlomo Sternberg

2121407 Semigroups.

Documents

2121407 Semigroups. - Harvard Universitypeople.math.harvard.edu/~shlomo/212a/07.pdf · The power series expansion of etA when A is bounded.Unbounded operators, their resolvents and